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# Scott, Jean , and warren are all building wooden models for an archite

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Director
Joined: 20 Feb 2015
Posts: 767
Concentration: Strategy, General Management
Scott, Jean , and warren are all building wooden models for an archite  [#permalink]

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Updated on: 03 Jul 2019, 04:42
1
1
00:00

Difficulty:

15% (low)

Question Stats:

90% (01:35) correct 10% (00:49) wrong based on 21 sessions

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Scott, Jean , and warren are all building wooden models for an architectural presentation at noon tomorrow. If their individual probabilities of finishing on time are x, $$\frac{1}{3}$$ , and $$\frac{1}{7}$$ , respectively , then what is the probability that warren will finish on time but Jean and Scott will not ?

a. $$\frac{(21x-38)}{21}$$

b. $$\frac{12x}{21}$$

c. $$\frac{(2-2x)}{21}$$

d. $$\frac{(2-4x)}{21}$$

e. $$\frac{(4x-56)}{21}$$

Originally posted by CounterSniper on 30 Sep 2018, 00:41.
Last edited by SajjadAhmad on 03 Jul 2019, 04:42, edited 1 time in total.
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Concentration: Finance, Marketing
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Re: Scott, Jean , and warren are all building wooden models for an archite  [#permalink]

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30 Sep 2018, 00:52
1
All are independent activities.

The probability of finishing the task on time by individuals

Scott = x

Jean = 1/3

Warren = 1/7

The probability of not finishing the task on time.

Scott = 1-x

jean = 1-1/3 = 2/3

Warren = 1-1/7 = 6/7

Now,
Warren finishes on time, But Scott and Jean Don't.

the probability that Warren will finish on time but Jean and Scott will not = (1-x)*2/3*1/7 = $$\frac{(2-2x)}{21}$$

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Re: Scott, Jean , and warren are all building wooden models for an archite   [#permalink] 30 Sep 2018, 00:52
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